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0px"><i class="fa fa-fw"></i><a href="../886/hybridization.html">Hybridization</a></label></li></ul></div> <button id="show-periodic-table" class="btn btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Further covalent bonding and structure <a href="#" class="mark-page-favorite pull-right" data-pid="885" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../357/bonding-and-structure.html">Bonding and Structure</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Further covalent bonding and structure</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60'</span> </ol> <article id="main-article"> <p> <img alt="" src="../../images/test-images/screenshot-2020-03-15-at-15.36.28.png" style="width: 160px; height: 86px; float: left;">There is quite a lot of varied content here. Most of it links to, and develops ideas of, models that we encountered in the <strong>covalent bonding</strong> and <strong>covalent structure </strong>sections (4.2 and 4.3) of the course, but also to the atomic structure; <strong>electron configuration</strong> section (2.2). It is worth ensuring that you are comfortable with the content in those sections before you begin your revision of this part of the course.</p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p><strong>a sigma (σ) bond*</strong>, <strong>a pi (π) bond*</strong>, <strong>VSEPR theory*</strong>, formal charge, valence electrons, expanded octet, resonance structures, delocalised electrons, conjugated system.</p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1,3" data-mode="Normal" data-topics="609" data-subject-id="7" data-n-flashcards="9" style="text-align:center">Show flashcards</a></div><hr> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>The Bohr atom</p> </div> </div> <div class="panel-body"> <div>A reminder from <strong>electron configuration</strong> (section 2.2) that our Lewis (dot-cross) diagrams are based on the Bohr model of the atom.</div> <div> </div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/283707909"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Understanding resonance</p> </div> </div> <div class="panel-body"> <div>A reminder from <strong>covalent structure</strong> (section 4.3) of how we explain resonance in terms of Lewis diagrams.</div> <div> </div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/407161843"></iframe></div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Formal Charge</p> </div> </div> <div class="panel-body"> <div>Understanding how to assign formal charge to atoms in Lewis structures.</div> <div> </div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/442426155"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <p> The revision cards contain all of the essential content:</p> <div id="carousel-188" class="dynamic-gallery carousel slide" data-id="188"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/7-188/screenshot-2020-07-27-at-164508.png" data-fancybox="gallery-188" title="" data-caption=""><img alt="" src="../../../std-galleries/7-188/screenshot-2020-07-27-at-164508.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-188/screenshot-2020-07-27-at-164526.png" 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data-stats="7-139-885"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Draw a Lewis (dot-cross) diagram for sulfur hexafluoride (SF<sub>6</sub>).</p><p>What is the shape around the central atom and what are the likely bond angles?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Hexahedral; 60° and 90°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Hexahedral; 90° and 180°</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Octahedral; 90° and 180°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Octahedral; 60° and 90°</span></label> </p></div><div class="q-explanation"><p>Sulfur hexafluoride is <strong>octahedral </strong>in shape, with bond angles of 90° and 180°. The name 'octahedral' comes from the <strong>number of faces </strong>of the shape (rather than the number of vertices/corners). As you may be able to see from the diagram below, the shape is two square-based pyramids base-to-base on top of each other making an octahedron.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/sf6.png" style="width: 432px; height: 222px;"><img alt="" height="223" src="../../images/chemical-bonding-and-structure/octahedron(1).png" width="223"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Draw a Lewis (dot-cross) diagram for xenon tetrafluoride (XeF<sub>4</sub>).</p><p>What is the shape around the central atom and what are the likely bond angles?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Tetrahedral; 109.5°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Tetrahedral; 60° and 90°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Octahedral; 90° and 180°</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Square planar; 90° and 180°</span></label> </p></div><div class="q-explanation"><p>Xenon tetrafluoride is <strong>square planar </strong>in shape, with bond angles of 90° and 180°.</p><p>The shape of the <strong>electron domains</strong> is octahedral, but with two vertices missing (the lone pairs) the bonds form the shape of a square. Imagine the octahedron shown below without the top and bottom vertices.</p><p><img alt="" height="202" src="../../images/chemical-bonding-and-structure/xef4.png" width="481"><img alt="" height="195" src="../../images/chemical-bonding-and-structure/octahedron(1).png" width="195"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Draw a Lewis (dot-cross) diagram for sulfur tetrafluoride (SF<sub>4</sub>).</p><p>What is the shape around the central atom and what are the likely bond angles?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Tetrahedral; 109.5°</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Seesaw (sawhorse); 90°, 120° and 180°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Octahedral; 90° and 180°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Square planar; 90° and 180°</span></label> </p></div><div class="q-explanation"><p>Sulfur tetrafluoride is <strong>seesaw (sawhorse) </strong>in shape, with bond angles of 90°, 120° and 180°.</p><p>The shape of the <strong>electron domains</strong> is trigonal bi-pyramidal, but with one equatorial (rather than axial) vertex missing (the lone pair) the bonds form the shape of a seesaw. Imagine the shape shown below with one equatorial vertex missing.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/sf4.png" style="width: 469px; height: 220px;"><img alt="" height="215" src="../../images/chemical-bonding-and-structure/hexhedron.png" width="215"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Draw a Lewis (dot-cross) diagram for iodine trifluoride (IF<sub>3</sub>).</p><p>What is the shape around the central atom and what are the likely bond angles?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>T–shaped; 90° and 180°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Trigonal planar; 120°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Trigonal bipyramidal; 90°, 120° and 180°</span></label> </p><p><label class="radio"> <input type="radio"> <span>Trigonal pyramidal; 107°</span></label> </p></div><div class="q-explanation"><p>Iodine trifluoride is <strong>T–shaped</strong>, with bond angles of 90° and 180°.</p><p>The shape of the <strong>electron domains</strong> is trigonal bi-pyramidal, but with two equatorial (rather than axial) vertices missing (the lone pairs) the bonds form the shape of a 'T'. Imagine the shape shown below with two equatorial vertices missing.</p><p><img alt="" height="207" src="../../images/chemical-bonding-and-structure/if3.png" width="482"><img alt="" height="208" src="../../images/chemical-bonding-and-structure/hexhedron.png" width="208"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following describe/s a sigma (σ) bond?</p><p><strong>1: </strong> The bond is formed by the end-to-end overlap of two atomic orbitals.</p><p><strong>2:</strong> The electrons in the bond lie mostly above and below the plane of the inter-nuclear axis.</p><p><strong>3:</strong> A sigma bond consists of two electrons .</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p></div><div class="q-explanation">A sigma bond is a molecular bonding orbital, and forms from the overlap of two atomic orbitals end-to-end (these can be s or p orbitals). A sigma bond does consist of two electrons. The electrons (or electron density) lie <strong>between the two nuclei</strong> (not above and below). 1 and 3 only is therefore the correct answer. The diagram below shows the formation of a sigma bond from 2 p orbitals:<img alt="" src="../../images/chemical-bonding-and-structure/sigma.png" style="width: 431px; height: 96px;"></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following describe/s a pi (π) bond?</p><p><strong>1: </strong> The bond is formed by the end-to-end overlap of two atomic orbitals.</p><p><strong>2:</strong> The electrons in the bond lie mostly above and below the plane of the inter-nuclear axis.</p><p><strong>3:</strong> A pi bond consists of two electrons .</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p></div><div class="q-explanation">A pi bond is a molecular bonding orbital, and forms from the overlap of two <strong>p </strong>(note p orbitals only) <strong>atomic orbitals side-on </strong>(not end-to-end) . A pi bond does consist of two electrons. The electrons (or electron density) lie mostly above and below the plane of the nuclei. 2 and 3 only is therefore the correct answer. The diagram below shows the formation of a pi bond from 2 p orbitals:<img alt="" src="../../images/chemical-bonding-and-structure/pi.png" style="width: 432px; height: 160px;"></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>How many sigma (σ) and pi (π) bonds in total are there in ethyne (H–C≡C–H)?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>3 sigma and 3 pi</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3 sigma and 2 pi </span></label> </p><p><label class="radio"> <input type="radio"> <span>1 sigma and 2 pi</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 sigma and 3 pi</span></label> </p></div><div class="q-explanation"><p>Every bond in the diagram is one pair of electrons. Every sigma bond and every pi bond is one pair of electrons, so there must be a total of 5 bonds.</p><p>A single bond is just a sigma bond.</p><p>A double bond is one sigma bond and one pi bond.</p><p>A triple bond is one sigma and two pi bonds.</p><p>There are therefore<em> 3 sigma bonds and 2 pi bonds </em>present in ethyne.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>When the bonding in a chemical species cannot be adequately represented by one single Lewis diagram, resonance structures may be used. When the resonance structures for a molecule of ozone, O<sub>3</sub>, are drawn, what are the <strong>formal charges</strong> on each of the oxygen atoms?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>+1: –1: –1</span></label> </p><p><label class="radio"> <input type="radio"> <span>0: 0: 0</span></label> </p><p><label class="radio"> <input type="radio"> <span>–2: –2: –2</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>0: +1: –1</span></label> </p></div><div class="q-explanation"><p>Formal charge = <strong>normal valence electrons</strong> – (non-bonding electrons + (½ × bonding electrons))</p><p>This can be confusing (and appear in different formats): formal charge works out the number of electrons an atom 'has' compared to how many it 'should have'.</p><p>The <strong>normal valence electrons</strong> is equal to the number of electrons in an atom's valence shell when in a 'normal' (ground or isolated) state. E.g oxygen in group VI (16) has 6 electrons in its valence shell.</p><p>(non-bonding electrons + (½ × bonding electrons)) are the electrons that the atom <strong>actually has in its current state</strong>.</p><p>The electrons an atom 'has' (all of the non-bonding electrons and half of the bonding electrons are 'owned' by that atom) are being taken away from the electrons it 'should have'. Atoms that have a formal charge of zero are in a 'normal' (electronically stable) state those that have a +ve formal charge are electron deficient; –ve, electron rich.</p><p>The oxygen atoms have FC of (left-to-right first resonance structure below):</p><p>FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0</p><p>FC = 6 – (2 + (½ × 6)) = 6 – (5) = +1</p><p>FC = 6 – (6 + (½ × 2)) = 6 – (7) = –1</p><p>(or vica versa) respectively.</p><p><img alt="" src="../../images/chemical-bonding-and-structure/ozone-lewis.png" style="width: 545px; height: 196px;"></p><p><img alt="" height="169" src="../../images/chemical-bonding-and-structure/ozone.png" width="546"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Sulfur dioxide can be represented by several different Lewis diagrams, two of which are shown below:</p><p style="text-align: center;"><img alt="" src="../../images/chemical-bonding-and-structure/formal-charge(1).png" style="width: 444px; height: 135px;"></p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td> </td><td style="text-align: center;"><strong>Lewis diagram A</strong></td><td style="text-align: center;"><strong>Lewis Diagram B</strong></td><td> </td></tr></tbody></table><p>What are the <strong>formal charges</strong> (FC) on each of the sulfur atoms in diagram A and B respectively, and which Lewis diagram is the most electronically stable representation of SO<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>FC 0 (sulfur A); FC +1 (sulfur B); most stable diagram is B</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>FC –4 (sulfur A); FC 0 (sulfur B); most stable diagram is A</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input class="c" type="radio"> <span>FC 0 (sulfur A); FC +1 (sulfur B); most stable diagram is A</span></label> </p><p><label class="radio" style=" float: left; margin-right: 40px; "> <input type="radio"> <span>FC –4 (sulfur A); FC 0 (sulfur B); most stable diagram is B</span></label> </p></div><div class="q-explanation"><p><strong>The atoms of Lewis diagram A have FC of:</strong></p><p>Sulfur FC = 6 – (2 + (½ × 8)) = 6 – (6) = 0</p><p>Oxygen (both the same) FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0</p><p><strong>The atoms of Lewis diagram B have FC of:</strong></p><p>Sulfur FC = 6 – (2 + (½ × 6)) = 6 – (5) = +1</p><p>Oxygen (double bonded) FC = 6 – (4 + (½ × 4)) = 6 – (6) = 0</p><p>Oxygen (single bonded) FC = 6 – (6 + (½ × 2)) = 6 – (7) = –1</p><p><strong>The first Lewis diagram, A, </strong>(in which the sulfur has expanded its octet)<strong> is therefore the most stable since all atoms in the structure have FC of zero.</strong></p><p>'FC 0 (sulfur A); FC +1 (sulfur B); most stable diagram is A' is the correct answer.</p><p>Remember:</p><p>Formal charge = <strong>normal valence electrons</strong> – (non-bonding electrons + (½ × bonding electrons))</p><p>This can be confusing (and appear in different formats): formal charge works out the number of electrons an atom 'has' compared to how many it 'should have'.</p><p>The <strong>normal valence electrons</strong> is equal to the number of electrons in an atom's valence shell when in a 'normal' (ground or isolated) state. E.g sulfur in group VI (16) has 6 electrons in its valence shell.</p><p>(non-bonding electrons + (½ × bonding electrons)) are the electrons that the atom <strong>actually has in its current state</strong>.</p><p>The electrons an atom 'has' (all of the non-bonding electrons and half of the bonding electrons are 'owned' by that atom) are being taken away from the electrons it 'should have'. Atoms that have a formal charge of zero are in a 'normal' (electronically stable) state those that have a +ve formal charge are electron deficient; –ve, electron rich.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>Oxygen molecules can be broken up into oxygen atoms:</p><p>O<sub>2</sub> → 2O</p><p>And oxygen atoms can then react with oxygen molecules to form ozone:</p><p>O<sub>2</sub> + O → O<sub>3</sub></p><p>Ozone can also dissociate back to an oxygen molecule and an oxygen atom :</p><p>O<sub>3</sub> → O<sub>2</sub> + O</p><p>Light from which region of the electromagnetic spectrum is required to dissociate oxygen and to dissociate ozone respectively?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Visible; ultra-violet</span></label> </p><p><label class="radio"> <input type="radio"> <span>Ultra-violet; infra-red</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Ultra-violet; ultra-violet</span></label> </p><p><label class="radio"> <input type="radio"> <span>Ultra-violet; visible</span></label> </p></div><div class="q-explanation"><p>Ultra-violet light is needed to dissociate both oxygen and ozone. The wavelength of light needed to dissociate oxygen is lower (the light is therefore higher in energy) than ozone. This is because the double bond in oxygen is stronger than the '1½' bonds in ozone (normally shown with the resonance structures as alternating double/single bonds).</p><p><img alt="" height="155" src="../../images/chemical-bonding-and-structure/ozone.png" width="501"></p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>Chlorofluorocarbons (CFCs) and oxides of nitrogen, NO<sub>x</sub> can catalyse the decomposition of ozone in the upper atmosphere.</p><p>Possible two-step mechanisms are given below for a chlorine radical (from CFCs) and for NO:</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;">O<sub>3</sub> + Cl → ClO + O<sub>2</sub></td><td style="text-align: center;">O<sub>3</sub> + NO → NO<sub>2</sub> + O<sub>2</sub></td></tr><tr><td style="text-align: center;">ClO + O<sub>3</sub> → Cl + 2O<sub>2</sub></td><td style="text-align: center;">NO<sub>2</sub> + O<sub>3</sub> → NO + 2O<sub>2</sub></td></tr></tbody></table><p>How many ozone molecules can<strong> one mole </strong>of chlorine radicals, and <strong>one mole </strong>of nitrogen monoxide molecules respectively, decompose?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>two moles; one mole</span></label> </p><p><label class="radio"> <input type="radio"> <span>two moles; two moles</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Many thousands; many thousands</span></label> </p><p><label class="radio"> <input type="radio"> <span>one mole; two moles</span></label> </p></div><div class="q-explanation"><p>The chlorine radicals and nitrogen monoxide are both acting as catalysts in these reactions. Thus, despite the mechanisms both showing a 1:2 molar ratio overall of catalyst to ozone molecules, both catalysts are re-produced at the end of the reaction, and are not 'used up', so they can go on decomposing many thousands of ozone molecules before they come to the end of their lifetimes.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">12</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Calculator question:</em> Given that the bond energy of an O=O bond is 498 kJ mol<sup>–1</sup>, calculate the wavelength of light (in metres) required to break a single O=O bond.</p><p>Use <em>E</em> = <em>h</em>ν and <span class="math-tex">\(c = νλ\)</span> from the data book.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>3.99×10<sup>−28</sup>m</span></label> </p><p><label class="radio"> <input type="radio"> <span>3.99×10<sup>−25</sup>m</span></label> </p><p><label class="radio"> <input type="radio"> 2.40×10<sup>−10</sup>m<span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2.40×10<sup>−7</sup>m</span></label> </p></div><div class="q-explanation"><p><em>E</em> = <em>h</em>ν (energy = Planck constant × frequency) is given in the data book.</p><p>Also in the data book is <span class="math-tex">\(c = νλ\)</span> (speed of light = frequency × wavelength)</p><p>Rearrangement of <span class="math-tex">\(c = νλ\)</span> will give <span class="math-tex">\(ν = {c \over λ}\)</span> (frequency = speed of light / wavelength).</p><p>So using <em>E</em> = <em>h</em>ν and <span class="math-tex">\(ν = {c \over λ}\)</span> by substituting frequency from the second expression into the first expression gives <span class="math-tex">\(E = {hc \over λ}\)</span>.</p><p>Energy is given for <strong>one mole</strong> of electrons in the question, so 498 kJ mol<sup>–1</sup> must be divided by Avogadro's number (<em>L</em>) to give the value for one bond.</p><p>498/6.02×10<sup>23</sup>=8.27...×10<sup>−22</sup> kJ which is 8.27...×10<sup>−19</sup> J (must be in Joules as Planck's constant is in <em>Joule seconds</em>, Js)</p><p>Thus, putting in all the known values, <span class="math-tex">\(E = {hc \over λ}\)</span> becomes:</p><p> <span class="math-tex">\({8.2724252×10^{-19}} = {(6.63×10^{-34})(3.00×10^{8}) \over λ}\)</span></p><p>Rearranging:</p><p><span class="math-tex">\({λ} = {(6.63×10^{-34})(3.00×10^{8}) \over 8.2724252×10^{-19}}\)</span></p><p>Wavelength is therefore: 2.40×10<sup>−7</sup> m (240nm)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button 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